$\sum\limits_{n=1}^{\infty } \dfrac{(x+3)^n}{n} $ What is the interval of convergence of the series? Choose 1 answer: Choose 1 answer: (Choice A) A $-5 \le x<-1$ (Choice B) B $-5 < x<-1$ (Choice C) C $-4 \le x<-2$ (Choice D) D $-4 < x<-2$
Solution: We use the ratio test. For $x\neq -3$, let $a_n= \dfrac{(x+3)^n}{n}$. $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|= \left| {x+3}\right| $ The series converges when $\left| {x+3}\right| <1$, which is when $-4<x<-2$. Now let's check the endpoints, $x=-4$ and $x=-2$. Letting $x=-4$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(-4+3)^n}{n} &=\sum\limits_{n=1}^{\infty } \dfrac{(-1)^n}{n} \end{aligned}$ This is the alternating harmonic series, which is known to converge. Letting $x=-2$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(-2+3)^n}{n} &=\sum\limits_{n=1}^{\infty } \dfrac{1}{n} \end{aligned}$ This is the harmonic series, which is known to diverge. In conclusion, the interval of convergence is $-4 \le x<-2$.